15HD_OJ题——Rightmost Digit

/*
 * Copyright (c) 2014, 烟台大学计算机学院
 * All rights reserved.
 * 文件名称：test.cpp
 * 作    者：李晓凯
 * 完成日期：2015年 5 月 24 日
 * 版 本 号：v1.0
 *
 * 问题描述：
 * 输入描述：
 * 程序输出：

 */

Problem Description

Given a positive integer N, you should output the most right digit of N^N.

Input

The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains a single positive integer N(1<=N<=1,000,000,000).

Output

For each test case, you should output the rightmost digit of N^N.

Sample Input

234

Sample Output

76

Hint

In the first case, 3 * 3 * 3 = 27, so the rightmost digit is 7.
In the second case, 4 * 4 * 4 * 4 = 256, so the rightmost digit is 6.

代码：

（1）

#include <iostream>using namespace std;int main(){    int n,T,m,a;    cin>>T;    while (T--)    {        cin>>n;        a=n%10;        if (a==0||a==1||a==5||a==6||a==9)            m=a;        else if (a==2)        {            if(n%4==0)                m=6;            else                m=4;        }        else if (a==3)        {            if (n%4==1)                m=3;            else                m=7;        }        else if (a==4)        {            m=6;        }        else if (a==7)        {            if (n%4==1)                m=7;            else                m=3;        }        else        {            if (n%4==0)                m=6;            else                m=4;        }        cout<<m<<endl;    }    return 0;}

（2）

#include <iostream>using namespace std;int main(){    int T,n,i,a,k;    cin>>T;    while(T--)    {        cin>>n;        a=n%10;        k=1;        for(i=0;i<=(n-1)%4;i++)            k=k*a;        cout<<k%10<<endl;    }    return 0;}